3.672 \(\int e^{-2 \tanh ^{-1}(a x)} (c-\frac{c}{a^2 x^2})^2 \, dx\)

Optimal. Leaf size=40 \[ \frac{c^2}{a^3 x^2}-\frac{c^2}{3 a^4 x^3}+\frac{2 c^2 \log (x)}{a}+c^2 (-x) \]

[Out]

-c^2/(3*a^4*x^3) + c^2/(a^3*x^2) - c^2*x + (2*c^2*Log[x])/a

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Rubi [A]  time = 0.116405, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6157, 6150, 75} \[ \frac{c^2}{a^3 x^2}-\frac{c^2}{3 a^4 x^3}+\frac{2 c^2 \log (x)}{a}+c^2 (-x) \]

Antiderivative was successfully verified.

[In]

Int[(c - c/(a^2*x^2))^2/E^(2*ArcTanh[a*x]),x]

[Out]

-c^2/(3*a^4*x^3) + c^2/(a^3*x^2) - c^2*x + (2*c^2*Log[x])/a

Rule 6157

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[d^p, Int[(u*(1 - a^2*x^
2)^p*E^(n*ArcTanh[a*x]))/x^(2*p), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rule 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rubi steps

\begin{align*} \int e^{-2 \tanh ^{-1}(a x)} \left (c-\frac{c}{a^2 x^2}\right )^2 \, dx &=\frac{c^2 \int \frac{e^{-2 \tanh ^{-1}(a x)} \left (1-a^2 x^2\right )^2}{x^4} \, dx}{a^4}\\ &=\frac{c^2 \int \frac{(1-a x)^3 (1+a x)}{x^4} \, dx}{a^4}\\ &=\frac{c^2 \int \left (-a^4+\frac{1}{x^4}-\frac{2 a}{x^3}+\frac{2 a^3}{x}\right ) \, dx}{a^4}\\ &=-\frac{c^2}{3 a^4 x^3}+\frac{c^2}{a^3 x^2}-c^2 x+\frac{2 c^2 \log (x)}{a}\\ \end{align*}

Mathematica [A]  time = 0.015661, size = 40, normalized size = 1. \[ \frac{c^2}{a^3 x^2}-\frac{c^2}{3 a^4 x^3}+\frac{2 c^2 \log (x)}{a}+c^2 (-x) \]

Antiderivative was successfully verified.

[In]

Integrate[(c - c/(a^2*x^2))^2/E^(2*ArcTanh[a*x]),x]

[Out]

-c^2/(3*a^4*x^3) + c^2/(a^3*x^2) - c^2*x + (2*c^2*Log[x])/a

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Maple [A]  time = 0.035, size = 39, normalized size = 1. \begin{align*} -{\frac{{c}^{2}}{3\,{a}^{4}{x}^{3}}}+{\frac{{c}^{2}}{{x}^{2}{a}^{3}}}-x{c}^{2}+2\,{\frac{{c}^{2}\ln \left ( x \right ) }{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c/a^2/x^2)^2/(a*x+1)^2*(-a^2*x^2+1),x)

[Out]

-1/3*c^2/a^4/x^3+c^2/x^2/a^3-x*c^2+2*c^2*ln(x)/a

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Maxima [A]  time = 0.959533, size = 51, normalized size = 1.27 \begin{align*} -c^{2} x + \frac{2 \, c^{2} \log \left (x\right )}{a} + \frac{3 \, a c^{2} x - c^{2}}{3 \, a^{4} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a^2/x^2)^2/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="maxima")

[Out]

-c^2*x + 2*c^2*log(x)/a + 1/3*(3*a*c^2*x - c^2)/(a^4*x^3)

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Fricas [A]  time = 1.73587, size = 99, normalized size = 2.48 \begin{align*} -\frac{3 \, a^{4} c^{2} x^{4} - 6 \, a^{3} c^{2} x^{3} \log \left (x\right ) - 3 \, a c^{2} x + c^{2}}{3 \, a^{4} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a^2/x^2)^2/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="fricas")

[Out]

-1/3*(3*a^4*c^2*x^4 - 6*a^3*c^2*x^3*log(x) - 3*a*c^2*x + c^2)/(a^4*x^3)

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Sympy [A]  time = 1.93487, size = 39, normalized size = 0.98 \begin{align*} \frac{- a^{4} c^{2} x + 2 a^{3} c^{2} \log{\left (x \right )} + \frac{3 a c^{2} x - c^{2}}{3 x^{3}}}{a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a**2/x**2)**2/(a*x+1)**2*(-a**2*x**2+1),x)

[Out]

(-a**4*c**2*x + 2*a**3*c**2*log(x) + (3*a*c**2*x - c**2)/(3*x**3))/a**4

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Giac [B]  time = 1.34723, size = 151, normalized size = 3.78 \begin{align*} -\frac{2 \, c^{2} \log \left (\frac{{\left | a x + 1 \right |}}{{\left (a x + 1\right )}^{2}{\left | a \right |}}\right )}{a} + \frac{2 \, c^{2} \log \left ({\left | -\frac{1}{a x + 1} + 1 \right |}\right )}{a} + \frac{{\left (3 \, c^{2} - \frac{5 \, c^{2}}{a x + 1} - \frac{3 \, c^{2}}{{\left (a x + 1\right )}^{2}} + \frac{6 \, c^{2}}{{\left (a x + 1\right )}^{3}}\right )}{\left (a x + 1\right )}}{3 \, a{\left (\frac{1}{a x + 1} - 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c/a^2/x^2)^2/(a*x+1)^2*(-a^2*x^2+1),x, algorithm="giac")

[Out]

-2*c^2*log(abs(a*x + 1)/((a*x + 1)^2*abs(a)))/a + 2*c^2*log(abs(-1/(a*x + 1) + 1))/a + 1/3*(3*c^2 - 5*c^2/(a*x
+ 1) - 3*c^2/(a*x + 1)^2 + 6*c^2/(a*x + 1)^3)*(a*x + 1)/(a*(1/(a*x + 1) - 1)^3)